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One mole of a diatomic.gas undergoes a p...

One mole of a diatomic.gas undergoes a process `p=(p_(0))/(1+((V)/(V_(0))^(3))`, where `p_(0)` and `V_(0)` are constants . The translational kinetic energy of the gas when V=`V_(0)` is given by

A

`(3P_(0)V_(0))/(2)`

B

`P_(0)V_(0)`

C

`(3P_(0)V_(0))/(4)`

D

`(5P_(0)V_(0))/(2)`

Text Solution

Verified by Experts

`P=(P_(0))/(1+((V)/(V_(0)))^(3))=(P_(0))/(2)`
`T=(P_(0)V_(0))/(2R)`
`:. ` Translational kinetic energy is equal to `(3)/(2)RT=(3R)/(2)(P_(0)V_(0))/(2R)=(3P_(0)V_(0))/(4)`
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